MCQ
Given :
$E^o_{Fe^{3+} /Fe} = -0.036\,V,$ $E^o _{Fe^{2+} /Fe} = -0.439\,V$
The value of standard electrode potential for the change,
$Fe^{3+} (aq) + e^- \rightarrow Fe^{2+} (aq)$ will be ........ $V$.
- A$0.385$
- ✓$0.770$
- C$-0.270$
- D$-0.072$
✓
Answer
Correct option: B.
$0.770$
b
$F e^{3+}+3 e^{-} \rightarrow F e, E^{o}_{F e^{3} 4 / F e}$
$F e^{3+}+3 e^{-} \rightarrow F e, E^{o}_{F e^{3} 4 / F e}$
$=-0.036 V \ldots(i)$
$F e^{2+}+2 e^{-} \rightarrow F e, E^{\circ}_{F e^{2} / / F e}$
$=-0.439 V \ldots(i i)$
we have to calculate
$F e^{3+}+e^{-} \rightarrow F e^{2+}, \Delta G=?$
To obtain this equation subtract equ $( ii )$ from $( i )$ we get
$F e^{3+}+e^{-} \rightarrow F e^{2+} \ldots(i i i)$
As we know that $\Delta G=-n F E$
Thus for reaction ( iii)
$\Delta G=\Delta G_{1}-\Delta G$
$-n F E^{\circ}=-n F E_{1}-\left(-n F E_{2}\right)$
$-n F E^{\circ}=n F E_{2}-n F E_{1}$
$-1 F E^{\circ}=2 \times 0.439 F-3 \times 0.036 F$
$- F E^{\circ}=0.770 F$
$\therefore \quad E^{\circ}=-0.770 V$
$O^{-}>F^{-}>N a^{+}>M g^{++}>A l^{3+}$
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